Question

The angle between the straight lines $x-1=\frac{2y+3}{3}=\frac{z+5}{2}$ and $x-3r+2; y=-2r-1; z=2,$ where $r$ is a parameter, is

• A. $\frac{\pi}{4}$
• B. $\cos ^{-1}\left(\frac{-3}{\sqrt{182}}\right)$
• C. $\sin ^{-1}\left(\frac{-3}{\sqrt{182}}\right)$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (D)

$\frac{\pi}{2}$

Answer 2

Given equation of lines are,
$x-1=\frac{2 y+3}{3}=\frac{z+5}{2}$
and $x=3 r+2 ; y=-2 r-1 ; z=2$
$\Rightarrow \frac{x-1}{1}=\frac{y+\frac{3}{2}}{\frac{3}{2}}=\frac{z+5}{2}$
and $\frac{x-2}{3}=r ; \frac{y+1}{-2}=r ; \frac{z-2}{0}=r$
$\Rightarrow \frac{x-1}{1}=\frac{y+\frac{3}{2}}{\frac{3}{2}}=\frac{z+5}{2}$
and $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$
DR's of lines Ist and IInd lines are $\left(1, \frac{3}{2}, 2\right)$ and $(3,-2,0)$
$\therefore$ The angle between two straight lines is
$\cos \,\theta =\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{1 \times 3-\frac{3}{2} \times 2+2 \times 0}{\sqrt{1^{2}+\left(\frac{3}{2}\right)^{2}+2^{2}} \sqrt{3^{2}+2^{2}+0^{2}}}$
$=\frac{3-3+0}{\sqrt{1+\frac{9}{4}+4} \sqrt{9+4}}$
$=\frac{0}{\sqrt{\frac{29}{4}} \sqrt{13}}$
$\Rightarrow \cos \theta=0$
$\Rightarrow \theta=\frac{\pi}{2}$